# Proving that $I+A$ is invertable when $A$ is nilpotent: What intuition leads to a particular approach?

In an answer to this question, it has been suggested to consider the following: $$(I+A)(\sum_{j=0}^n(-A)^j)$$

Through a series of algebraic operations, it can be shown that $$\sum_{j=0}^n(-A)^j$$ is in fact the inverse of $$I+A$$.

How would we have known to multiply by $$\sum_{j=0}^n(-A)^j$$? If there isn't an identity or formula that would indicate such a multiplication is a reasonable avenue of inquiry, then how would we otherwise derive $$\sum_{j=0}^n(-A)^j$$?

• Just think that multiplying by some kind of sum of the powers of $\;A\;$ could help since $\;A^n=0\;$ for some natural $\;n\;$ ... For example, if $\;A^2=0\;$ , then $\;(I+A)(I-A)=I-A^2=I\;$ . This and, of course, the fact that $\;I\;$ commutes with any matrix (of the same order) makes things very simple. Jul 20, 2019 at 17:30
• To come up with the proof, it's best to start with a different (more structural) proof: Working in the quotient ring $\mathbb{Z}\left[A\right]/\left(I+A\right)$, we have $I \equiv -A \mod I+A$, so that $I$ is nilpotent modulo $I+A$ (since $A$ is nilpotent, and thus so is $-A$); but this means that $I^k \equiv 0 \mod I+A$ for some $k$, and therefore $I \equiv 0 \mod I+A$, and thus $I+A$ is invertible (since $1 = 0$ only holds in trivial rings). Now, unravel the use of quotient rings and congruences in this proof, extracting an explicit (or recursive) formula for the inverse. Jul 20, 2019 at 20:44

If $$z\in\mathbb C$$ and $$\lvert z\rvert<1$$, then$$\frac1{1+z}=1-z+z^2-z^3+\cdots$$In other words,$$(1+z)\left(1-z+z^2-z^3+\cdots\right)=1.$$This, together with the fact that $$A^n=0$$ if $$n\gg1$$, should make you think that it would be a good idea to try to prove that an inverse of $$\operatorname{Id}+A$$ is $$1-A+A^2-A^3+\cdots$$ (which is a finite sum, in this case).
I guess a natural way of thinking about this stems from the formula for a geometric series for real numbers: if $$|x| < 1$$, then \begin{align} \sum_{j=0}^\infty x^j = \dfrac{1}{1-x} \end{align} Now, replacing $$x$$ with $$-x$$ gives the formula \begin{align} (1+x)^{-1} = \dfrac{1}{1+x} = \sum_{j=0}^{\infty} (-x)^j \end{align} So, it might be natural to try this for matrices as well. And now if $$A$$ is nilpotent, then on the RHS, we only have a finite number of summands (in fact at most $$n$$ summands if $$A$$ is $$n \times n$$). Hence, one might expect that $$\sum_{j=0}^n (-A)^j$$ is the inverse matrix of $$I+A$$. Then, a simple computation verifies that it actually is.
Recall that $$x^{2n+1}+y^{2n+1}=(x+y)(x^{2n}-x^{2n-1}y+x^2y^{2n-2}-..+y^{2n})$$
Now, since $$I$$ and $$A$$ commute, the same formula holds for $$x=I$$ and $$y=A$$. Therefore, $$I+A^{2n+1}=(I+A)(I-A+A^2-A^{3}+...+A^{2n})$$ Now use the fact that $$A^n=0$$ implies $$A^{n}=A^{n+1}=...=A^{2n+1}=0$$